Mathematica - Analysis of a mass-damper-spring system

[Keesjanss]

Hello,

I am currently working on a problem, but at the moment I am stuck. I just don't know how to solve the problem so I hope someone can help me with it. This is the question

1a : Consider the standard mass-damper-spring system:
m y''+γ y'+k y=u

where the constants have the following values:
m=28
γ=3
k=13

First find the homogeneous solution to the differential equation using the DSolve command. This is also known as the transient solution. In other words, find the solution where u(t)=0. Use the initial values y(0)=1 and y^\[Prime](0)=1.

(* Use Set to assign your answer to the variable before this comment *)(* \
Your answer should be in the default output form produced by DSolve, \
which is a Rule within Lists.
The form should look similar to the following: *)

1b : What is the natural frequency Subscript[\[Omega], 0] of this system?


I hope someone can help me

Greets

 

[Keesjanss]

This is what I have so far

DSolve[28 y''[t] + 3 y'[t] + 13 y[t] == 0, y[t], t]

{{y[t] ->
E^(-3 t/56) C[2] Cos[(Sqrt[1447] t)/56] +
E^(-3 t/56) C[1] Sin[(Sqrt[1447] t)/56]}}

Now is my question, how do I lose the constants C[2] and C[1]. I think it has something to do with that y[0]=1 and y'[0]=1, but I don't know how I should insert those values in the equeation.

 

[Dale]

You can do it one of two ways. The first is to take your output function and use it to evaluate y[0]==1 and y'[0]==1. That is two equations which you can solve for C[1] and C[2].

The other way is to add those initial conditions to the equations in DSolve. So instead of just solving the differential equation, make a list like {diffeq, y[0]==1, y'[0]==1} or {diffeq, y[0]==y0, y'[0]==v0}

 

[Keesjanss]

Ok, but how do I exactly add those initial conditions in the equation

I have now : DSolve[28''y[t]+3y'[t]+13y[t]==0,y[t],t]

So where in the equation has the ''y[0]==1 andd y'[0]==1 be placed and which brackets do I have to use?

 

[Keesjanss]

Oh I think I got it correct now :

DSolve[{28 y''[t] + 3 y'[t] + 13 y[t] == 0, y[0] == 1, y'[0] == 1},
y[t], t]

and now there no constants anymore in the equation.

But now is the next question : What is the natural frequency w₀ of this system?

Someone who knows that?

 

[ganeshrk]

Your mass is 28; spring constant or stiffness is 13. Radian natural frequency is sqrt(k/m) = 0.681 radians/sec or 0.1084 Hertz (cycles/sec).

 

[AlephZero]

1b : What is the natural frequency Subscript[\[Omega], 0] of this system?

If you want the natural frequency of the damped system, think what the cos and sin terms in the solution mean!

Ganesrk found the natural frequency of the undamped system, which is a different value.

 

[ganeshrk]

damping ratio = c/(2sqrt(mk)) = 2/(2sqrt(28x13)) = 0.0786 | 7.8% critical damping

damped radian freq = unadamped radian freq x sqrt(1 - damping ratio^2) = 0.679

The amplitude of motion of the mass is logarithmically decreasing and it is a function of the damping ratio.